Question

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A raindrop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms-1?
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Answer 1

ANSWER

- 0.162 Joules

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Answer 2

★Given:-

• Radius of raindrop = 2 mm
• It falls from a height of 500 m above the ground.
• It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original heigh.
• It attains its maximum (terminal) speed, and moves with uniform speed thereafter.

★To find:-

• The work done by the gravitational force on the drop in the first and second half of its journey?
• The work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms-1?

★Solution:-

We have,

◖Radius of drop=2mm

                       =2×10⁻³m

◖Density of water = 10³ Kg/m³

As the Raindrop is a sphere,

◖The volume of the drop will be 4/3πr³.

Using the formula,

✦Mass = Density × volume

Mass of the rain drop:-

➜M = ρV

➜4/3πr³ × 10³ Kg

➜4/3 × 3.14 × (2× 10^-3)³ × 10³ Kg

➜3.35 × 10⁻⁵ Kg

Given,rain drop falls from the height of 500m.

Distance travelled in 1st half journey:-

➜500/2

➜250 m

Using the formula,

✦Work done = F.d = mgh

Work done by Gravitational force on the drop in 1st half journey:-

Work done = mg × distance travelled

➜3.35 × 10⁻⁵ × 9.8 × 250 Joule

➜820 × 10^-4 Joule

➜ 0.082 Joule

Work done by Gravitational force on the drop in 2nd half journey will be same as that of in the first journey as the distance travelled is same in both cases.

Therefore,

The work done by the gravitational force on the drop in the first and second half of its journey is 0.082J.

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To find the work done by resistive force:-

We know,

◖The total energy of the drop remains conserved during its motion.

◖During free fall the body has initial velocity 0.

◖Due to the presence of a resistive force, the drop hits the ground with a velocity of 10m/s.

Therefore,

• Initial velocity of rain drop = 0
• Final velocity of rain drop = 10m/s

Change in kinetic energy  = Final kinetic energy - initial kinetic energy

➜1/2mv² - 1/2mu²

➜1/2× 3.35 × 10⁻⁵ × (10)² - 0

➜1.675 × 10⁻³Joule

Total work done by Gravitational force on rain drop:-

➜0.082 + 0.082

➜2(0.082)

➜0.164 J

We know,

Change in kinetic energy = Workdone by the resistive force in the entire journey + total work done by Gravitational force on rain drop.

Putting values,

➜1.675 × 10⁻³ = Workdone by the resistive force + 0.164

➜Work done by resistive force = 1.675 × 10⁻³ - 0.164

➜ -0.162 joule

Hence,the work done by resistive force in the entire journey if its speed on reaching the ground is 10 m/s is -0.162J.

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